# example of changing variable

If one performs in the improper integral

 $\displaystyle I\;:=\;\int_{-\infty}^{\infty}\frac{e^{kx}}{1\!+\!e^{x}}\,dx% \qquad(0 (1)

the change of variable (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)

 $x\;=\;-\ln{t},\quad dx=-\frac{dt}{t},$

the new lower limit  becomes $\infty$ and the new upper limit 0; hence one obtains

 $I\;=\;-\int_{\infty}^{0}\frac{e^{-k\ln{t}}dt}{(1\!+\!e^{-\ln{t}})t}\;=\;\int_{% 0}^{\infty}\frac{t^{-k}}{t\!+\!1}\,dt.$

Thus one has recurred $I$ to the integral

 $\displaystyle\int_{0}^{\infty}\frac{x^{-k}}{x\!+\!1}\,dx,$ (2)

the value of which has been determined in the entry using residue theorem  near branch point  .  Accordingly, we may write the result

 $\int_{-\infty}^{\infty}\frac{e^{kx}}{1\!+\!e^{x}}\,dx\;=\;\frac{\pi}{\sin{\pi k% }}.$

Calculating the integral (1) directly is quite laborious:  one has to use Cauchy residue theorem to the integral

 $\oint_{c}\frac{e^{kz}}{1\!+\!e^{z}}\,dz$

about the perimetre $c$ of the rectangle

 $-a\,\leqq\,\mbox{Re}\,z\,\leqq\,a,\quad 0\,\leqq\,\mbox{Im}\,z\,\leqq\,2\pi$

and then to let  $a\to\infty$ (one cannot use the same half-disk as in determining the integral (2)).  As for using the method (http://planetmath.org/MethodsOfEvaluatingImproperIntegrals) of differentiation  under the integral sign or taking Laplace transform  with respect to $k$ yields a more complicated integral.

Title example of changing variable ExampleOfChangingVariable 2013-03-22 18:45:49 2013-03-22 18:45:49 pahio (2872) pahio (2872) 5 pahio (2872) Example msc 26A06 UsingResidueTheoremNearBranchPoint MethodsOfEvaluatingImproperIntegrals